Problem: $\overline{AB}$ = $3\sqrt{5}$ $\overline{BC} = {?}$ A C B 3\sqrt{5} ? $ \sin( \angle BAC ) = \frac{2\sqrt{5} }{5}, \cos( \angle BAC ) = \frac{ \sqrt{5}}{5}, \tan( \angle BAC ) = 2$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{3\sqrt{5}} $ $ \overline{BC}=3\sqrt{5} \cdot \sin( \angle BAC ) = 3\sqrt{5} \cdot \frac{2\sqrt{5} }{5} = 6$